An object is to be thrown through a window 3.25 m high a distance 8 m away. It is launched at an angle 75 degree above the horizontal.

The initial launch speed at the specified angle which will allow the projectile to pass through the window is 13.3 m/s.

What is the final speed of the object (that is, the speed at which it flies through the window)?
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But then the answer i got by using the formula Vf = Vo+at
= 13.3+ (-9.81)(2.85)
= -14.7 m/s
IS wrong!!What is my mistake??

oh...i'm so lost once again. I tought we should get the time By using:
y= yo+Voy*t-(1/2)gt^2 formula,
= 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2
If not can you explain me how to find the time and how to get the components Vfy and Vfx.